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Graph x^2/4 + y^2 = 1

Step by step, with the reason for every step · Ellipse · Precalculus

Step by step

The question

Question: find the center, the vertices and the foci of x² over 4 + y² = 1.

x24 + y2 = 1

1 Find the center

There are no brackets: just x² and y².

So the center is (0, 0).

center = (0, 0)

2 Find a and b

The bigger bottom is a²: 4.

It is under x.

So the long side goes across.

bigger bottom = a²: under x → wide

a² = 4, so a = 2: 2 times 2 is 4.

a is half the long side.

a2 = 4 → a = 2

The smaller bottom is b² = 1, so b = 1.

b is half the short side.

b2 = 1 → b = 1

3 Vertices

The vertices are the ends of the long side: go a = 2 from the center, left and right.

vertices = (−2, 0) and (2, 0)

Draw the grid.

Put the center on the grid.

The two vertices, 2 each way.

And b = 1 the other way: the ends of the short side.

A smooth oval through the four points.

4 Find the foci

The two foci sit inside, on the long side.

Their distance c from the center: c² = a² − b².

c2 = a2 − b2

Careful!

For an ellipse it is minus: c is shorter than a, so the foci stay inside.

Careful: not c² = 4 + 1 ✗

Now put the numbers in: a² = 4 and b² = 1.

c2 = 4 − 1

4 minus 1 is 3.

c2 = 3

So c is the square root of 3.

c = √3

3 is not a square number.

So we keep √3.

That is the exact answer.

It is about 1.7.

√3 ≈ 1.7

Go c from the center along the long side.

The foci are (−√3, 0) and (√3, 0).

foci = (−√3, 0) and (√3, 0)

Put the foci on the grid.

5 Check

Let's check with the end of the short side, (0, 1).

Join it to both foci.

Each line goes √3 across and 1 up: Pythagoras.

c² plus b² is 3 plus 1: 4.

The square root is 2.

distance = √3 + 1 = √4 = 2

The two distances add up to 4, the length of the long side.

That is true for every point of an ellipse.

Right!

2 + 2 = 4, long side 2 × 2 = 4

Let's look back.

First: find the center.

Then: find a and b.

Then: vertices.

Then: find the foci.

Last: check.

Answer
vertices = (−2, 0) and (2, 0)

Your turn

Try this one yourself, then check:

x29 + y2 = 1
Show the answer
V (−3, 0), (3, 0); F (−2√2, 0), (2√2, 0)

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